又有问题要问大牛们了 Is rs=rx-2? (1)r is an odd number (2)x=s+2 答案给的是B,解释是: restate the equation as rs=rs+2r-2;therefore.2=2r,and r=1. If r=1, then the eqution above can be simplified to s=x-2, which is the same statement as (2), which we know to be true. So (2) is sufficient 但我认为,问题问的是 rs=rx-2是否成立,即r(s-x)=-2是否成立。已知的条件只有一个x=s+2,故r(s-x)=-2r。如果r=1,则r(s-x)=-2r=-2,原来的等式rs=rx-2成立,如果r不等于1,则原来的等式不成立。